Question 17.4: Calculating Concentrations of Species in a Solution of a Dip...
Calculating Concentrations of Species in a Solution of a Diprotic Acid
Ascorbic acid (vitamin C) is a diprotic acid, \mathrm{H}_{2} \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6} (Figure 17.5). What is the \mathrm{pH} of a 0.10 \mathrm{M} solution? What is the concentration of ascorbate ion, \mathrm{C}_{6} \mathrm{H}_{6} \mathrm{O}_{6}{ }^{2-} ? The acid ionization constants are K_{a 1}=7.9 \times 10^{-5} and K_{a 2}=1.6 \times 10^{-12}.
PROBLEM STRATEGY
Diprotic acids have two ionization constants, one for the loss of each proton. To be exact, you should account for both reactions. However, K_{a 2} is so much smaller than K_{a 1} that the smaller amount of hydronium ion produced in the second reaction can be neglected. Follow the three steps for solving equilibrium problems.
Considering only the first ionization, set up a table of concentrations (starting, change, and equilibrium). Substitute these values into the equilibrium equation for K_{a 1}, solve for x=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right], and then solve for \mathrm{pH} (as in Example 17.2). Ascorbate ion is produced in the second ionization; its concentration, as we show, equals K_{a 2}.
CALCULATION of pH Abbreviate the formula for ascorbic acid as \mathrm{H}_{2} Asc. Hydronium ions are produced by two successive acid ionizations.
\begin{aligned} & \mathrm{H}_{2} \mathrm{Asc}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{HAsc}^{-}(a q) ; K_{a 1}=7.9 \times 10^{-5} \\ & \mathrm{HAsc}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\operatorname{Asc}^{2-}(a q) ; K_{a 2}=1.6 \times 10^{-12} \end{aligned}
If you let x be the amount of \mathrm{H}_{3} \mathrm{O}^{+}formed in the first ionization, you get the following results:
\begin{array}{cccc} \text { Concentration }(M) & \mathbf{H}_{2} \mathbf{A s c}(\boldsymbol{a q})+\mathbf{H}_{2} \mathbf{O}(\boldsymbol{l}) & \rightleftharpoons \mathbf{H}_{3} \mathbf{O}^{+}(\boldsymbol{a q})&+\mathbf{H A s c}^{-}(\boldsymbol{a q}) \\ \text { Starting } & 0.10 & \sim 0 & 0 \\ \text { Change } & -x & +x & +x \\ \text { Equilibrium } & 0.10-x & x & x \end{array}
Now substitute into the equilibrium-constant equation for the first ionization.
\begin{aligned} \frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{HAsc}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{Asc}\right]} & =K_{a 1} \\ \frac{x^{2}}{0.10-x} & =7.9 \times 10^{-5} \end{aligned}
Assuming x to be much smaller than 0.10 , you get
\frac{x^{2}}{0.10} \simeq 7.9 \times 10^{-5}
or
\begin{gathered} x^{2} \simeq 7.9 \times 10^{-5} \times 0.10 \\ x \simeq 2.8 \times 10^{-3}=0.0028 \end{gathered}
(Note that 0.10-x=0.10-0.0028=0.10, correct to two significant figures. Therefore, the assumption that 0.10-x \simeq 0.10 is correct.)
The hydronium-ion concentration is 0.0028 M, so
\mathrm{pH}=-\log \left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=-\log (0.0028)=\mathbf{2 . 5 5}
ASCORBATE-ION CONCENTRATION Ascorbate ion, \mathrm{Asc}^{2-}, which we will call y, is produced only in the second reaction. Assume the starting concentrations of \mathrm{H}_{3} \mathrm{O}^{+} and \mathrm{HAsc}^{-}for this reaction to be those from the first equilibrium. The amounts of each species in 1 \mathrm{~L} of solution are as follows:
\begin{array}{cccc} \text { Concentration }(M) & \mathbf{H A s c}^{-}(\boldsymbol{a q})+\mathbf{H}_{2} \mathbf{O}(l) & \rightleftharpoons \mathbf{H}_{3} \mathbf{O}^{+}(\boldsymbol{a q}) & +\mathbf{A s c}^{2-}(\boldsymbol{a q}) \\ \text { Starting } & 0.0028 & 0.0028 & 0 \\ \text { Change } & -y & +y & +y \\ \text { Equilibrium } & 0.0028-y & 0.0028+y & y \end{array}
Now substitute into the equilibrium-constant equation for the second ionization.
\begin{aligned} & \frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{Asc}^{2-}\right]}{\left[\mathrm{HAsc}^{-}\right]}=K_{a 2} \\ & \frac{(0.0028+y) y}{0.0028-y}=1.6 \times 10^{-12} \end{aligned}
This equation can be simplified if you assume that y is much smaller than 0.0028 . (Again, this assumes that the reaction occurs to only a small extent, as you expect from the magnitude of the equilibrium constant.) That is,
\begin{aligned} & 0.0028+y \simeq 0.0028 \\ & 0.0028-y \simeq 0.0028 \end{aligned}
Then the equilibrium equation reads
\frac{(0.0028) y}{0.0028} \simeq 1.6 \times 10^{-12}
Hence, y \simeq 1.6 \times 10^{-12}
(Note that 1.6 \times 10^{-12} is indeed much smaller than 0.0028 , as you assumed.) The concentration of ascorbate ion equals K_{a 2}, or 1.6 \times 10^{-12} M.